\(\int \frac {1}{(b x+c x^2)^2} \, dx\) [272]

Optimal result

Integrand size = 11, antiderivative size = 43 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=-\frac {b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac {2 c \log (x)}{b^3}+\frac {2 c \log (b+c x)}{b^3} \]

[Out]

(-2*c*x-b)/b^2/(c*x^2+b*x)-2*c*ln(x)/b^3+2*c*ln(c*x+b)/b^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {628, 629} \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=-\frac {2 c \log (x)}{b^3}+\frac {2 c \log (b+c x)}{b^3}-\frac {b+2 c x}{b^2 \left (b x+c x^2\right )} \]

[In]

Int[(b*x + c*x^2)^(-2),x]

[Out]

-((b + 2*c*x)/(b^2*(b*x + c*x^2))) - (2*c*Log[x])/b^3 + (2*c*Log[b + c*x])/b^3

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 629

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[Log[x]/b, x] - Simp[Log[RemoveContent[b + c*x, x]]/b,
x] /; FreeQ[{b, c}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac {(2 c) \int \frac {1}{b x+c x^2} \, dx}{b^2} \\ & = -\frac {b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac {2 c \log (x)}{b^3}+\frac {2 c \log (b+c x)}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=-\frac {b \left (\frac {1}{x}+\frac {c}{b+c x}\right )+2 c \log (x)-2 c \log (b+c x)}{b^3} \]

[In]

Integrate[(b*x + c*x^2)^(-2),x]

[Out]

-((b*(x^(-1) + c/(b + c*x)) + 2*c*Log[x] - 2*c*Log[b + c*x])/b^3)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {1}{b^{2} x}-\frac {2 c \ln \left (x \right )}{b^{3}}-\frac {c}{b^{2} \left (c x +b \right )}+\frac {2 c \ln \left (c x +b \right )}{b^{3}}\)	\(43\)
risch	\(\frac {-\frac {2 c x}{b^{2}}-\frac {1}{b}}{x \left (c x +b \right )}+\frac {2 c \ln \left (-c x -b \right )}{b^{3}}-\frac {2 c \ln \left (x \right )}{b^{3}}\)	\(49\)
norman	\(\frac {\frac {2 c^{2} x^{2}}{b^{3}}-\frac {1}{b}}{x \left (c x +b \right )}-\frac {2 c \ln \left (x \right )}{b^{3}}+\frac {2 c \ln \left (c x +b \right )}{b^{3}}\)	\(50\)
parallelrisch	\(-\frac {2 \ln \left (x \right ) x^{2} c^{2}-2 \ln \left (c x +b \right ) x^{2} c^{2}+2 \ln \left (x \right ) x b c -2 \ln \left (c x +b \right ) x b c -2 c^{2} x^{2}+b^{2}}{b^{3} x \left (c x +b \right )}\)	\(70\)

[In]

int(1/(c*x^2+b*x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/b^2/x-2*c*ln(x)/b^3-c/b^2/(c*x+b)+2*c*ln(c*x+b)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=-\frac {2 \, b c x + b^{2} - 2 \, {\left (c^{2} x^{2} + b c x\right )} \log \left (c x + b\right ) + 2 \, {\left (c^{2} x^{2} + b c x\right )} \log \left (x\right )}{b^{3} c x^{2} + b^{4} x} \]

[In]

integrate(1/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

-(2*b*c*x + b^2 - 2*(c^2*x^2 + b*c*x)*log(c*x + b) + 2*(c^2*x^2 + b*c*x)*log(x))/(b^3*c*x^2 + b^4*x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=\frac {- b - 2 c x}{b^{3} x + b^{2} c x^{2}} + \frac {2 c \left (- \log {\left (x \right )} + \log {\left (\frac {b}{c} + x \right )}\right )}{b^{3}} \]

[In]

integrate(1/(c*x**2+b*x)**2,x)

[Out]

(-b - 2*c*x)/(b**3*x + b**2*c*x**2) + 2*c*(-log(x) + log(b/c + x))/b**3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=-\frac {2 \, c x + b}{b^{2} c x^{2} + b^{3} x} + \frac {2 \, c \log \left (c x + b\right )}{b^{3}} - \frac {2 \, c \log \left (x\right )}{b^{3}} \]

[In]

integrate(1/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

-(2*c*x + b)/(b^2*c*x^2 + b^3*x) + 2*c*log(c*x + b)/b^3 - 2*c*log(x)/b^3

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=\frac {2 \, c \log \left ({\left | c x + b \right |}\right )}{b^{3}} - \frac {2 \, c \log \left ({\left | x \right |}\right )}{b^{3}} - \frac {2 \, c x + b}{{\left (c x^{2} + b x\right )} b^{2}} \]

[In]

integrate(1/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

2*c*log(abs(c*x + b))/b^3 - 2*c*log(abs(x))/b^3 - (2*c*x + b)/((c*x^2 + b*x)*b^2)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (b x+c x^2\right )^2} \, dx=\frac {4\,c\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b^3}-\frac {\frac {1}{b}+\frac {2\,c\,x}{b^2}}{c\,x^2+b\,x} \]

[In]

int(1/(b*x + c*x^2)^2,x)

[Out]

(4*c*atanh((2*c*x)/b + 1))/b^3 - (1/b + (2*c*x)/b^2)/(b*x + c*x^2)